A 10 kg box slides down a long, frictionless incline of angle 30°. It starts fro

Question

A 10 kg box slides down a long, frictionless incline of angle 30°. It starts from rest at time t = 0 at the top of the incline at a height of 24 m above ground.

(a) What is the original potential energy of the box relative to the ground?
J

(b) From Newton's laws, find the distance the box travels in 1 s and its speed at t = 1 s.
m
m/s

(c) Find the potential energy and the kinetic energy of the box at t = 1 s.
J (kinetic energy)
J (potential energy)

(d) Find the kinetic energy and the speed of the box just as it reaches the bottom of the incline.
J
m/s

Explanation / Answer

part a )

Ui = mgh

Ui = 10*9.8*24 = 2352 J

part b )

a = gsin30

x = 1/2*a*t^2

x = 1/2 * gsin30 * 1

x = 1/2 * 9.8 * .5 * 1

x = 2.45 m

speed = at

gsin30 = 4.9 m/s

part c )

K = 1/2 * m*v^2

K = 1/2 * 10 * (4.9)^2

K = 120.05 J

U = Ui-K

U = 2352 - 120.05 = 2231.95 J

part d )

at bottom potential energy will totally convert into kinetic energy

K = 2352 J

speed = sqrt(2K/m)

v = 21.69 m/s

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